Change of Measure and Expectations Under Girsanov’s Theorem

# On Girsanov Theorem to switch from Risk-Neutral to Stock Numeraire


# On Girsanov Theorem to switch from Risk-Neutral to Stock Numeraire












Summary: long-story cut short, the question is asking for what types of functions $f(.)$, the Cameron-Martin-Girsanov theorem can be used as follows:

$$ \mathbb{E}^{\mathbb{P}^2}[f(W_t)]=\mathbb{E}^{\mathbb{P}^1}\left[\frac{d\mathbb{P}^2}{d\mathbb{P}^1}f(W_t)\right] $$

Long story: the Radon-Nikodym when changing from risk-neutral to Stock measure is:

$$\frac{dN^{S}}{dN^{Q}}=\frac{N^{Q}_{t_0}}{N^{Q}_{t}} \frac{N^{S}_{t}}{N^{S}_{t_0}}=\frac{1}{e^{rt}}\frac{S_t}{S_{t_0}}=e^{-0.5\sigma^2t+\sigma W_t}$$

The following type of calc is often seen in finance:

$$\mathbb{E}^{N^S}\left[S_t \right]=\mathbb{E}^{N^Q}\left[S_t^Q \frac{dN^{S}}{dN^{Q}} \right]=\\=\mathbb{E}[S_t^Q*e^{-0.5\sigma^2t+\sigma W_t}]=\\=S_0e^{rt-0.5\sigma^2t+\sigma W_t}*e^{-0.5\sigma^2t+\sigma W_t}=\\=S_0e^{rt+\sigma^2t}$$

The CMG theorem tells us that the Radon-Nikodym derivative $e^{-0.5\sigma^2t+\sigma W_t}$ can be applied to $W_t$ directly to modify it's drift and to create some new measure under which $W_t$ will no longer be a Standard Brownian motion. If we step-away from finance and denote the measure under which $W_t$ is standard Brownian as $\mathbb{P}^1$, the new measure under which $W_t$ is a Brownian with a drift as $\mathbb{P}^2$, and the radon-nikodym as $\frac{d\mathbb{P}^2}{d\mathbb{P}^1}$, we can write:

$$\mathbb{P}^2(W_t<a):=\mathbb{E}^{\mathbb{P}^1}\left[\frac{d\mathbb{P}^2}{d\mathbb{P}^1} * I_{\{W_t<a\}} \right] =\mathbb{E}^{\mathbb{P}^1}\left[e^{-0.5\sigma^2t+\sigma W_t} * I_{\{W_t<a\}} \right] $$

The above is basically the definition of $\mathbb{P^2}$ via the implicit definition of the Radon-Nikodym derivative. An extension of the above definition is that:

$$ \mathbb{E}^{\mathbb{P}^2}[W_t]=\mathbb{E}^{\mathbb{P}^1}\left[\frac{d\mathbb{P}^2}{d\mathbb{P}^1}W_t\right] $$

Question: in our finance case of stock, the stock price process is actually a function of $W_t$, so we could write $S_t=f(W_t)$. In the equation $\mathbb{E}^{N^S}\left[S_t \right]=\mathbb{E}^{N^Q}\left[S_t^Q \frac{dN^{S}}{dN^{Q}} \right]$, we are actually using the fact that:

$$ \mathbb{E}^{\mathbb{P}^2}[f(W_t)]=\mathbb{E}^{\mathbb{P}^1}\left[\frac{d\mathbb{P}^2}{d\mathbb{P}^1}f(W_t)\right] $$

Is there an easy way to prove that we can do that? Obviously it does work as shown in the case of the stock price process above, because it produces the correct result. But for what $f(.)$ does the result hold? I am sure there must be some restrictions on the types of functions $f(.)$ for which the result holds true.

## Answer by ir7 (score 3, accepted)

https://quant.stackexchange.com/a/55864

(I might not be answering your question, but I feel this clarification is needed.)

A random variable $X$ of $(\Omega, \mathcal{F})$ is a $\mathcal{F}$-measurable function $X : \Omega → \mathbf{R}$. So, $X$ depends on $\Omega$ and $\mathcal{F}$, but does not depend on the probability measure put on $(\Omega, \mathcal{F})$. It is the distribution of $X$ that depends on the measure.

Given $P1$ and $P_2$ probability measures on $(\Omega, \mathcal{F})$, where $P_2$ is $P_1$-absolutely continuous on $\mathcal{F}$ and $$ L = \frac{dP_2}{dP_1} $$ is the Radon-Nicodym derivative ($\mathcal{F}$-measurable, $\mathcal{P_1}$-integrable), we have: $$X\in L^1(\Omega, P_2) \iff XL\in L^1(\Omega, P_1).$$ In that case, we then have: $$ \mathbf{E}^{P_2}[X] = \mathbf{E}^{P_1}[XL] $$

or in its integral form:

$$ \int_\Omega X dP_2 = \int_\Omega X \frac{dP_2}{dP_1} dP_1 $$

(Note that there is no need to introduce notation $X^{P_2}$ competing with $X$.)

For your question:

$$ \mathbf{E}^{P_2}[f(W_t)] = \mathbf{E}^{P_2}[f(W_t^\theta -\int_0^t \theta_u du)] $$

if $P_2$ is the Girsanov measure built from process $\theta$ and $W_t^\theta = W_t +\int_0^t \theta_u du$ is the induced Brownian motion under $P_2$ ($W_t$ is a Brownian motion under $P_1$). One can compute the expectation under $P_2$. Or go back to $P_1$ as you said:

$$ \mathbf{E}^{P_2}[f(W_t)] = \mathbf{E}^{P_1}\left[f(W_t)\frac{dP_2}{dP_1} \right]. $$

In your case $\theta_t = \sigma$ and

$$ \frac{dP_2}{dP_1} =\exp\left(-\frac{\sigma^2}{2} t + \sigma W_t \right). $$